Problem: Solve for $n$: $2^n\cdot 4^n=64^{n-36}$.
Explanation: Since $4=2^2$, $4^n=2^{2n}$.  Since $64=2^6$, $64^{n-36}=2^{6(n-36)}$.  Thus,

$$2^{n+2n}=2^{6(n-36)}\Rightarrow 3n=6n-216$$

So $3n=216\Rightarrow n=\boxed{72}$.